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This is an excerpt from Zorich's book. I have some issues with understanding the last paragraph of it. This is what I understood so far:

Suppose we have polynomial with complex coefficients $P(z)$. We would like to find roots and their multiplicities.

Suppose $q(z):=\text{gcd}(P(z),P'(z)).$ Suppose we know factorization of $q(z)$, i.e. $q(z)=(z-z_1)^{\alpha_1}\dots (z-z_p)^{\alpha_p}$ then somehow he deduces that roots of $P(z)$ are also $z_1,\dots,z_p$ with multiplicities $\alpha_1+1,\dots, \alpha_p+1$.

Am I right that h is doing exactly this? If yes how to prove it? I am bit confused.

RFZ
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  • That's correct. Suppose $(z-z_k)$ is a factor of $q = gcd(P,P')$, then it is by definition a factor of $P$. That it has multiplicity $1$ greater follows from the quoted corollary. Note that this only allows you to find roots of multiplicity $\gt 1$, if any such exist at all. – dxiv Jul 02 '21 at 00:25
  • @dxiv, I think it is not enough. Also how you prove that any root of $P$ is also a root $q$? – RFZ Jul 02 '21 at 00:28
  • Not any root of $P$ is a root of $q$, and no such claim is made. The proposition is only about roots of $P$ of multiplicity $\gt 1$. – dxiv Jul 02 '21 at 00:30
  • @dxiv, I don't think you're right because he is trying to find a polynomial $p(z)$ whose roots are the same as $P(z)$ but have multiplicity $1$. – RFZ Jul 02 '21 at 00:31
  • @dxiv, Can you give a separate answer with details? I'll appreciate it! – RFZ Jul 02 '21 at 00:32
  • Try it with a simple example e.g. $P(z)=z^2(z-1)$, then $q(z)=z$ and $p(z) = P(z)/q(z)=z(z-1)$. Guess I am not quite sure where you see the problem. – dxiv Jul 02 '21 at 00:34

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Let $q$ be the monic $\gcd(P, P^\prime)$. Since $q$ is a monic divisor of $P$, then it must be of the form $$q(z) = (z - z_1)^{\alpha_1} \cdots (z - z_p)^{\alpha_p}$$ where each $\alpha_i$ is nonnegative and at most the multiplicity of the root $z_i$ in $P$.

There is an inverse statement to the corollary which is not stated that if $z_i$ is a root of $P$ of multiplicity $1$, then it is not a root of $P^\prime$.

Since $q$ is also a divisor of $P’$, there are two cases:

  • The number $\alpha_i = 0$ and thus $z_i$ is not a root of $P^\prime$.
  • The number $\alpha_i$ is positive, and thus $z_i$ is a common root of $P$ and $P^\prime$.

In the first case, the contrapositive to the corollary tells us that the multiplicity of $z_i$ in $P$ is $1$. In the second case, the inverse statement to the corollary shows that the multiplicity of $z_i$ in $P$ is greater than $1$.

In both cases, the maximal power of $(z - z_i)$ that divides $P$ and $P^\prime$ is $(z - z_i)^{k_i - 1}$, and $\alpha_i = k_i - 1$ where $k_i$ is the multiplicity of the root $z_i$ in $P$.

shoteyes
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  • Thanks a lot for your reply! Let me ask you a question regarding your first line: do you apply Fundamental Theorem of Algebra to $P(z)$ or $q(z)$? From your wording it seems that it has been applied to $P(z)$, right? – RFZ Jul 02 '21 at 01:04
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    @ZFR I’m assuming that $P$ can be factored into linear terms, yes. From that, it follows that $q$ can be factored into linear terms as well! – shoteyes Jul 02 '21 at 01:07